f(x) ={|x|, x≤12−x, x>1 , then f[f(x)] is equal to

ff(x) ={f|x|, x≤1f(2−x), x>1 f(|x|) ={|x|, x≤1 and |x|≤12−|x|, x≤1 and |x|>1={|x|, −1≤x≤12−|x|, x 1 and 2−x≤12−(2−x), x>1 and 2−x>1 (impossible) =|2−x|, x>1∴ f[f(x)] ={2−|x|, x 1

f(x) ={|x|, x≤12−x, x>1 , then f[f(x)] is equal to

ff(x) ={f|x|, x≤1f(2−x), x>1 f(|x|) ={|x|, x≤1 and |x|≤12−|x|, x≤1 and |x|>1={|x|, −1≤x≤12−|x|, x 1 and 2−x≤12−(2−x), x>1 and 2−x>1 (impossible) =|2−x|, x>1∴ f[f(x)] ={2−|x|, x 1

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$f (x)$ $= {\begin{matrix} | x |, x \leq 1 \\ 2 - x, x > 1 \end{matrix}$ , then $f [f (x)]$ is equal to

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None of these

${\begin{matrix} 2 - | x |, x \leq - 1 \\ | x |, - 1 \leq x \leq 1 \\ | 2 - x |, x > 1 \end{matrix}$

${\begin{matrix} | 2 - x |, x < - 1 \\ | x |, - 1 \leq x \leq 1 \\ 2 - | x |, x > 1 \end{matrix}$

${\begin{matrix} | x |, x \leq - 1 \\ 2 - | x |, - 1 \leq x \leq 1 \\ | 2 - x |, x > 1 \end{matrix}$

answer is A.

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Detailed Solution

$f (f (x))$ $= {\begin{matrix} f | x |, x \leq 1 \\ f (2 - x), x > 1 \end{matrix}$

$f (| x |)$ $= {\begin{matrix} | x |, x \leq 1 and | x | \leq 1 \\ 2 - | x |, x \leq 1 and | x | > 1 \end{matrix} = {\begin{matrix} | x |, - 1 \leq x \leq 1 \\ 2 - | x |, x < - 1 \end{matrix}$